Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $z \neq 0$. $r = \dfrac{-8z + 32}{z - 10} \div \dfrac{9z - 90}{z^2 - 20z + 100} $
Dividing by an expression is the same as multiplying by its inverse. $r = \dfrac{-8z + 32}{z - 10} \times \dfrac{z^2 - 20z + 100}{9z - 90} $ First factor the quadratic. $r = \dfrac{-8z + 32}{z - 10} \times \dfrac{(z - 10)(z - 10)}{9z - 90} $ Then factor out any other terms. $r = \dfrac{-8(z - 4)}{z - 10} \times \dfrac{(z - 10)(z - 10)}{9(z - 10)} $ Then multiply the two numerators and multiply the two denominators. $r = \dfrac{ -8(z - 4) \times (z - 10)(z - 10) } { (z - 10) \times 9(z - 10) } $ $r = \dfrac{ -8(z - 4)(z - 10)(z - 10)}{ 9(z - 10)(z - 10)} $ Notice that $(z - 10)$ appears twice in both the numerator and denominator so we can cancel them. $r = \dfrac{ -8(z - 4)\cancel{(z - 10)}(z - 10)}{ 9\cancel{(z - 10)}(z - 10)} $ We are dividing by $z - 10$ , so $z - 10 \neq 0$ Therefore, $z \neq 10$ $r = \dfrac{ -8(z - 4)\cancel{(z - 10)}\cancel{(z - 10)}}{ 9\cancel{(z - 10)}\cancel{(z - 10)}} $ We are dividing by $z - 10$ , so $z - 10 \neq 0$ Therefore, $z \neq 10$ $r = \dfrac{-8(z - 4)}{9} ; \space z \neq 10 $